Pressure in Fluids and its Transmission - Questions & Answers

Explained, Step by Step

EXERCISE-4

(A) MULTIPLE CHOICE TYPE :
(Choose the correct answer from the options given below).

1. The thrust exerted by a body placed on a surface is :
(a) less than the weight of the body
(b) more than the weight of the body
(c) equal to the weight of the body
(d) independent of the weight of the body
Answer: (c) equal to the weight of the body.
Step 1: Understand that thrust is defined as the total perpendicular force applied by an object on a surface.
Step 2: When an object simply rests on a horizontal surface, the only downward force it exerts is its weight due to gravity.
Step 3: Therefore, the thrust is exactly equal to the weight of the body.

2. The pressure exerted on a surface depends on :
(a) the nature of the surface on which the thrust is applied
(b) the area on which the thrust is applied
(c) the magnitude of the thrust applied
(d) both (b) and (c)
Answer: (d) both (b) and (c).
Step 1: Recall the mathematical formula for pressure: Pressure = Thrust / Area.
Step 2: Look at the numerator; pressure depends directly on the magnitude of the thrust.
Step 3: Look at the denominator; pressure depends inversely on the area over which it is applied.
Step 4: Therefore, pressure depends on both the thrust and the area.

3. Pressure and thrust are :
(a) vector quantities
(b) scalar quantities
(c) scalar and vector quantities respectively
(d) vector and scalar quantities respectively
Answer: (c) scalar and vector quantities respectively.
Step 1: Recall that pressure acts in all directions in a fluid and does not have a single specific direction, making it a scalar quantity.
Step 2: Recall that thrust is a force acting specifically in a perpendicular direction, making it a vector quantity.
Step 3: Therefore, the correct sequence is scalar (for pressure) and vector (for thrust).

4. The C.G.S. unit of pressure is :
(a) Pa
(b) Nm^-2
(c) dyne cm^-2
(d) dyne cm²
Answer: (c) dyne cm^-2.
Step 1: Identify the C.G.S. unit for thrust (force), which is the dyne.
Step 2: Identify the C.G.S. unit for area, which is square centimeters (cm²).
Step 3: Since Pressure = Thrust / Area, the unit becomes dyne / cm², or dyne cm^-2.

5. One bar is equal to :
(a) 10³ Nm^-2
(b) 10⁴ N m^-2
(c) 10⁵ Nm^-2
(d) 10⁶ Nm²
Answer: (c) 10⁵ Nm^-2.
Step 1: Recall standard unit conversions taught in the chapter.
Step 2: The unit 'bar' is a meteorological unit for pressure.
Step 3: By definition, 1 bar is exactly equal to 100,000 pascals, which is 10⁵ N m^-2.

6. 1 atm of pressure is equal to :
(a) 0.076 m of Hg
(b) 760 torr
(c) 76 torr
(d) 7.6 mm of Hg
Answer: (b) 760 torr.
Step 1: Recall standard atmospheric pressure values.
Step 2: 1 atmosphere (atm) supports a mercury column of 76 cm or 760 mm.
Step 3: Since 1 mm of Hg is called 1 torr, 760 mm of Hg equals 760 torr.

7. The pressure on a surface is reduced by :
(a) using high thrust
(b) increasing the area of surface
(c) decreasing the area of surface
(d) none of these
Answer: (b) increasing the area of surface.
Step 1: Write down the formula: Pressure = Thrust / Area.
Step 2: Notice that pressure and area are inversely proportional.
Step 3: To make the pressure smaller (reduce it), you must make the denominator (area) larger.

8. Cutting tools have either sharp or pointed edges so that a ........... thrust may cause a ........ pressure at the edges and cutting can be done with less effort.
(a) high, high
(b) small, small
(c) high, small
(d) small, high
Answer: (d) small, high.
Step 1: A sharp edge means the surface area is extremely small.
Step 2: Because Pressure = Thrust / Area, a tiny area creates a very large (high) pressure even with a small push.
Step 3: Therefore, a small thrust easily causes a high pressure to cut through objects.

9. The pressure inside a liquid of density ρ at a depth h is :
(a) hρg
(b) h / ρg
(c) hρ / g
(d) hρ
Answer: (a) hρg.
Step 1: Recall the derivation for liquid pressure.
Step 2: Pressure is caused by the weight of the liquid column above.
Step 3: The standard formula derived in the textbook is Pressure = depth × density × gravity = hρg.

10. Out of the following, which one is correct about the law of liquid pressure ?
(a) Pressure is different in all directions about a point inside the liquid.
(b) In a stationary liquid, pressure is different at all points on a horizontal plane.
(c) Inside the liquid, pressure decreases with an increase in depth from its free surface.
(d) It increases with an increase in the density of the liquid.
Answer: (d) It increases with an increase in the density of the liquid.
Step 1: Review the laws of liquid pressure step by step.
Step 2: Pressure is actually the same in all directions at a point (eliminates a).
Step 3: Pressure is the same at all points on a horizontal plane (eliminates b).
Step 4: Pressure increases (not decreases) with depth (eliminates c).
Step 5: Looking at P = hρg, pressure is directly proportional to density, making (d) the correct statement.

11. The pressure P₁ at a certain depth in river water and P₂ at the same depth in sea water are related as :
(a) P₁ > P₂
(b) P₁ = P₂
(c) P₁ < P₂
(d) P₁ - P₂ = atmospheric pressure
Answer: (c) P₁ < P₂.
Step 1: Identify that depth (h) and gravity (g) are the same in both cases.
Step 2: Recall that sea water is salty and has a higher density than fresh river water.
Step 3: Since Pressure depends on density (P = hρg), higher density means higher pressure.
Step 4: Therefore, pressure in river water (P₁) is less than pressure in sea water (P₂).

12. The pressure P₁ at the top of a dam and P₂ at a depth h from the top inside water (density ρ) are related as :
(a) P₁ > P₂
(b) P₁ = P₂
(c) P₁ - P₂ = hρg
(d) P₂ - P₁ = hρg
Answer: (d) P₂ - P₁ = hρg.
Step 1: Let P₁ be the atmospheric pressure acting at the top surface of the dam.
Step 2: At depth h, the total pressure P₂ includes the surface pressure plus the liquid pressure: P₂ = P₁ + hρg.
Step 3: Rearrange the equation by moving P₁ to the other side.
Step 4: This gives P₂ - P₁ = hρg.

13. The wall of a dam is made thicker at the bottom because :
(a) the pressure exerted by a liquid remains the same with its depth.
(b) the pressure exerted by a liquid decreases with its depth.
(c) the pressure exerted by a liquid increases with its depth.
(d) of safety from outer forces.
Answer: (c) the pressure exerted by a liquid increases with its depth.
Step 1: Understand that water in a dam is stored to a great height.
Step 2: According to the formula P = hρg, pressure increases as depth (h) increases.
Step 3: At the very bottom of the dam, the depth is maximum, so the water exerts massive crushing pressure.
Step 4: The walls must be built thicker at the base to withstand this extreme pressure without breaking.

14. Hydraulic machines work on the principle of :
(a) Newton's first law
(b) Newton's third law
(c) Pascal's law
(d) both (a) and (b)
Answer: (c) Pascal's law.
Step 1: Recall how hydraulic machines transfer force using liquids.
Step 2: The core concept is that applying pressure to confined fluid transmits it equally in all directions.
Step 3: This specific transmission of fluid pressure is defined by Pascal's law.

15. Hydraulic machines act like a :
(a) force subtractor
(b) force reducer
(c) force multiplier
(d) both (a) and (c)
Answer: (c) force multiplier.
Step 1: Visualize a hydraulic machine with a small pump piston and a large lifting piston.
Step 2: A small effort force applied on the small piston creates pressure.
Step 3: This pressure acts on the much larger area of the second piston to create a massively larger output force.
Step 4: Because a small force is turned into a large force, it multiplies the force.

16. A rectangular cement block has length, breadth and height of 60 cm, 30 cm and 15 cm respectively. Identify the correct position of the block and pressure exerted by it on the ground.
(a) Minimum pressure is exerted when breadth and height form the base.
(b) Maximum pressure is exerted when breadth and height form the base.
(c) Maximum pressure is exerted when length and breadth form the base.
(d) Minimum pressure is exerted when length and height form the base.
Answer: (b) Maximum pressure is exerted when breadth and height form the base.
Step 1: Note that the weight (thrust) of the block is always the same regardless of how it is placed.
Step 2: Pressure is maximum when the base area touching the ground is the smallest.
Step 3: Calculate area 1: Length × Breadth = 60 × 30 = 1800 cm².
Step 4: Calculate area 2: Length × Height = 60 × 15 = 900 cm².
Step 5: Calculate area 3: Breadth × Height = 30 × 15 = 450 cm².
Step 6: Since Breadth and Height give the smallest area (450 cm²), they will exert the maximum pressure.

17. The incorrect statement from the following is :
(a) The pressure exerted by the liquid filled in a vessel is same at all points.
(b) The pressure of liquid is same in all directions in a horizontal plane.
(c) The pressure of a liquid on a surface does not depend on the area of surface.
(d) The pressure of liquid at its free surface is zero.
Answer: (a) The pressure exerted by the liquid filled in a vessel is same at all points.
Step 1: Check statement (b): This is a true law of liquid pressure.
Step 2: Check statement (c): True, liquid pressure depends only on h, ρ, and g.
Step 3: Check statement (d): True, liquid column depth is zero at the surface.
Step 4: Check statement (a): False, because pressure changes depending on depth. Points deeper down have more pressure, so it is not the same at all points.

(B) VERY SHORT ANSWER TYPE :

1. Define the term thrust. State its S.I. unit.
Step 1: Thrust is defined as the force acting normally (perpendicularly) on a surface.
Step 2: Since thrust is a type of force, its S.I. unit is the newton (N).

2. (a) What physical quantity is measured in bar ?
(b) How is the unit bar related to the S.I. unit pascal ?
Step 1: (a) The unit 'bar' is used to measure the physical quantity known as pressure.
Step 2: (b) One bar is defined as exactly equal to 100,000 pascals.
Step 3: Therefore, the relationship is written as 1 bar = 10⁵ Pa.

3. Define one pascal (Pa), the S.I. unit of pressure.
Step 1: Start with the formula Pressure = Force / Area.
Step 2: Substitute 1 unit for force and 1 unit for area.
Step 3: One pascal is defined as the pressure exerted on a surface area of 1 square meter by a force of 1 newton acting normally on it.

4. State whether thrust is a scalar or vector ?
Step 1: Remember that thrust is a normal force.
Step 2: Force always has a specific direction (in this case, perpendicular to the surface).
Step 3: Because it has both magnitude and direction, it is a vector quantity.

5. State whether pressure is a scalar or vector ?
Step 1: Remember that fluid pressure acts equally in all directions at a given point.
Step 2: Because it doesn't have a single specific direction of action, it is a scalar quantity.

6. What is a fluid ?
Step 1: Think about states of matter that don't have a fixed shape and can spread out.
Step 2: A substance which has the tendency to flow is called a fluid.
Step 3: This category includes all liquids and gases.

7. Pressure at free surface of a water lake is P₁, while at a point at depth h below its free surface is P₂. (a) How are P₁ and P₂ related ? (b) Which is more P₁ or P₂ ?
Step 1: (a) Total pressure at a depth includes the surface pressure plus the liquid column pressure.
Step 2: The formula relating them is P₂ = P₁ + hρg.
Step 3: (b) Because you are adding the liquid pressure (hρg) to P₁, P₂ must be greater than P₁.

8. How does the liquid pressure on a diver change if :
(i) the diver moves to the greater depth, and
(ii) the diver moves horizontally ?
Step 1: (i) When moving to a greater depth, 'h' increases. According to P = hρg, the pressure will increase.
Step 2: (ii) When moving horizontally, depth 'h' stays exactly the same. Therefore, the pressure remains unchanged.

9. State Pascal's law of transmission of pressure.
Step 1: Focus on what happens when you press on an enclosed liquid.
Step 2: Pascal's law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

10. Name two applications of Pascal's law.
Step 1: Think of heavy machinery that lifts or presses things using fluids.
Step 2: Application one: The hydraulic press (used for compressing cotton bales).
Step 3: Application two: The hydraulic jack (used for lifting cars in service stations).

11. Complete the following sentences :
(a) Pressure at a depth h in a liquid of density ρ is hρg.
(b) Pressure is same in all directions about a point in a liquid.
(c) Pressure at all points at the same depth is same.
(d) Pressure at a point inside a liquid is directly proportional to its depth.
(e) Pressure of a liquid at a given depth is directly proportional to the density of liquid.

(C) SHORT ANSWER TYPE :

1. What is meant by pressure ? State its S.I. unit.
Step 1: Pressure describes the effect of a force spread over an area.
Step 2: It is defined as the thrust (normal force) acting per unit area of a surface.
Step 3: Its standard S.I. unit is the pascal (Pa), which is equal to 1 N m^-2.

2. Differentiate between thrust and pressure.
Step 1: Define both: Thrust is the total normal force acting on a surface, while pressure is the thrust acting per unit area.
Step 2: Note their mathematical nature: Thrust is a vector quantity, while pressure is a scalar quantity.
Step 3: Note their units: The SI unit of thrust is the newton (N), while the SI unit of pressure is the pascal (Pa).

3. How does the pressure exerted by a thrust depend on the area of surface on which it acts ? Explain with a suitable example.
Step 1: State the relationship based on the formula P = F / A.
Step 2: Pressure is inversely proportional to the area of the surface; a smaller area results in a larger pressure.
Step 3: Give an example: If you stand up on loose sand, your entire weight is concentrated on the small area of your feet, exerting high pressure and causing you to sink.
Step 4: If you lie down, your weight is spread over the large area of your whole body, reducing the pressure so you don't sink.

4. Why is the tip of an allpin made sharp ?
Step 1: Understand that an allpin needs to easily pierce materials like paper or cloth.
Step 2: Making the tip sharp makes its surface area extremely small.
Step 3: Because Pressure = Thrust / Area, applying a small force on this tiny area creates a huge pressure, easily driving the pin through the surface.

5. Explain the following :
(a) It is easier to cut with a sharp knife than with a blunt one.
(b) Sleepers are laid below the rails.
Step 1: (a) A sharp knife has a very thin, small edge area compared to a blunt one.
Step 2: For the same force applied by your hand, the small area creates much greater pressure, making cutting effortless.
Step 3: (b) Heavy trains exert massive downward thrust on tracks.
Step 4: Placing wide wooden or concrete sleepers underneath significantly increases the surface area.
Step 5: This spread-out area lowers the pressure on the soil, preventing the heavy rails from sinking into the ground.

6. What do you mean by the term fluid pressure ?
Step 1: Recognize that a fluid (liquid or gas) has weight and tends to flow.
Step 2: Because of this, it pushes against whatever is holding it.
Step 3: Fluid pressure is the pressure exerted by a fluid at all points and in all directions on the walls and bottom of its container.

7. How does the pressure exerted by a solid and a fluid differ ?
Step 1: A solid is rigid, so its weight only pushes in one direction: downwards against the surface supporting it.
Step 2: A fluid can move and spread out freely.
Step 3: Therefore, a fluid exerts pressure not just downwards on the base, but horizontally on the walls and equally in all directions inside the liquid.

8. State three factors on which the pressure at a point in a liquid depends.
Step 1: Look at the liquid pressure formula: P = hρg.
Step 2: Factor 1: The depth (h) of the point below the free surface of the liquid.
Step 3: Factor 2: The density (ρ) of the given liquid.
Step 4: Factor 3: The acceleration due to gravity (g) at that location.

9. Write an expression for the pressure at a point inside a liquid. Explain the meaning of the symbols used.
Step 1: Write the total pressure equation considering atmospheric pressure above the liquid.
Step 2: The expression is P = P₀ + hρg.
Step 3: 'P' stands for the total pressure at that depth.
Step 4: 'P₀' stands for atmospheric pressure on the free surface.
Step 5: 'h' is the vertical depth of the point from the surface.
Step 6: 'ρ' is the density of the liquid, and 'g' is acceleration due to gravity.

10. How does the pressure at a certain depth in sea water differ from that at the same depth in river water ? Explain your answer.
Step 1: Note that sea water contains dissolved salts, which makes it heavier per unit volume than fresh river water.
Step 2: Therefore, the density of sea water is greater than the density of river water.
Step 3: According to the formula P = hρg, higher density directly causes higher pressure at the same depth.
Step 4: Thus, the pressure at a certain depth in sea water is greater than at the same depth in river water.

11. Explain why a gas bubble released at the bottom of a lake grows in size as it rises to the surface of lake.
Step 1: When the bubble is at the bottom of the lake, it is under a large amount of pressure due to the deep water column above it.
Step 2: As the bubble rises up, the depth (h) of the liquid above it steadily decreases.
Step 3: This causes the surrounding liquid pressure on the bubble to decrease.
Step 4: According to Boyle's law for gases, a decrease in pressure causes a gas to expand.
Step 5: Because the pressure drops, the gas bubble's volume expands and it grows in size.

12. A dam has broader walls at the bottom than at the top. Explain.
Step 1: Remember the law of liquid pressure: Pressure is directly proportional to depth (P = hρg).
Step 2: At the top of the water level in the reservoir, the depth is zero, so the water pressure is zero.
Step 3: As you go deeper towards the bottom of the dam, the depth 'h' becomes massive, creating enormous water pressure pushing against the wall.
Step 4: Engineers build the dam wall thick and broad at the bottom to provide enough strength to safely hold back this massive pressure.

13. Why do sea divers need special protective suit ?
Step 1: Deep underwater, a diver experiences immense liquid pressure from the towering column of water above them.
Step 2: This external pressure is much higher than normal human blood pressure inside the body.
Step 3: Without protection, this extreme pressure could crush a diver's body.
Step 4: Divers wear suits made of rigid materials like glass-reinforced plastic or cast aluminum which maintain normal pressure inside the suit, protecting them.

14. State the laws of liquid pressure.
Step 1: First law: Inside a liquid, pressure increases with the increase in depth from its free surface.
Step 2: Second law: In a stationary liquid, pressure is the same at all points on a horizontal plane.
Step 3: Third law: Pressure is the same in all directions about a single point inside the liquid.
Step 4: Fourth law: Pressure at the same depth is different in different liquids; it increases as density increases.
Step 5: Fifth law: A liquid seeks its own level (it naturally settles to a flat surface).

15. A tall vertical cylinder filled with water is kept on a horizontal table top. Two small holes A and B are made on the wall of the cylinder, A near the middle and B just below the free surface of water. State and explain your observation.
Step 1: Observation: The water spurts out horizontally with a much greater force and lands further away from hole A, while it simply trickles out close to the cylinder from hole B.
Step 2: Explanation: Hole A is near the middle, meaning it has a significant depth of water above it.
Step 3: Hole B is just below the surface, so it has almost zero depth above it.
Step 4: Because liquid pressure increases with depth, the pressure driving the water out at A is much greater, causing it to shoot further.

16. Name and state the principle on which a hydraulic press works. Write one use of the hydraulic press.
Step 1: Name the principle: It works on Pascal's law of transmission of pressure.
Step 2: State the principle: It states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.
Step 3: State a use: It is used in industry for pressing heavy items like cotton bales, or squeezing oil out of seeds.

(D) LONG ANSWER TYPE :

1. Describe a simple experiment to demonstrate that a liquid enclosed in a vessel exerts pressure in all directions.
Step 1: Take a specialized glass flask that has several narrow tubes poking out from its sides and bottom.
Step 2: Fit the mouth of the flask with a tight, movable piston.
Step 3: Fill the flask and tubes completely with water until the liquid level is the same in all the narrow tubes.
Step 4: Press the piston down firmly to apply pressure to the enclosed water inside the flask.
Step 5: Observation: You will see the water level shoot up in all the narrow tubes simultaneously, reaching exactly the same height in each one.
Step 6: Conclusion: This proves that the single pressure applied at the top by the piston was transmitted equally sideways and downwards in all directions through the liquid.

2. Deduce an expression for the pressure at a depth inside a liquid.
Step 1: Imagine a container filled with a liquid of density 'ρ'.
Step 2: Imagine a flat, horizontal circular area 'A' at a depth 'h' below the liquid surface.
Step 3: The liquid directly above this area forms a virtual cylinder of height 'h' and base area 'A'.
Step 4: Calculate the volume of this liquid cylinder: Volume = Area × height = A × h.
Step 5: Calculate the mass of this liquid column: Mass = Volume × density = (A × h) × ρ.
Step 6: The thrust (force) on the bottom area is the weight of this liquid column: Thrust = Mass × gravity = A × h × ρ × g.
Step 7: Substitute this into the pressure formula: Pressure = Thrust / Area = (A × h × ρ × g) / A.
Step 8: Cancel out the 'A' from the numerator and denominator to get the final expression: P = hρg.

3. Explain the principle of a hydraulic machine. Name two devices which work on this principle.
Step 1: A hydraulic machine uses Pascal's law to function as a force multiplier.
Step 2: It consists of two connected cylinders filled with fluid: a narrow one with a small piston and a wide one with a large piston.
Step 3: When a small force (effort) is applied to the small piston, it creates pressure in the fluid.
Step 4: According to Pascal's law, this exact same pressure is transmitted through the fluid to the large piston.
Step 5: Because the second piston has a much larger area, the transmitted pressure multiplies into a much larger upward force (Force = Pressure × large Area).
Step 6: Two common devices using this are the hydraulic jack and hydraulic brakes.

4. The diagram in Fig. 4.12 below shows a device which makes use of the principle of transmission of pressure.
(a) Name the parts labelled by the letters X and Y.
(b) Describe what happens to the valves A and B and to the quantity of water in the two cylinders when the lever arm is moved down.
(c) State one use of the above device.
Step 1: (a) Recognizing the diagram of a hydraulic press, X is the large press plunger (or ram) and Y is the small pump plunger.
Step 2: (b) When the lever pushes plunger Y down, pressure in that cylinder spikes.
Step 3: This high pressure forces the one-way valve A to close so water can't go backwards.
Step 4: The same pressure forces valve B to open, allowing a quantity of water to transfer from the small cylinder into the large cylinder.
Step 5: (c) A major use for this device is lifting heavy vehicles or pressing cotton bales.

5. Draw a simple diagram of a hydraulic jack and explain its principle.
Step 1: (Diagram conceptually features a narrow cylinder with a small lever-operated piston, connected via a pipe with a valve to a much wider cylinder containing a large lifting platform, all filled with hydraulic fluid).
Step 2: Principle: It is built directly on Pascal's law of pressure transmission.
Step 3: You apply a small downward effort on the lever attached to the small piston.
Step 4: This pressurizes the confined fluid. The pressure travels undiminished to the wider cylinder.
Step 5: The larger area of the second cylinder turns this pressure into a massively multiplied upward thrust.
Step 6: This multiplied force is strong enough to easily lift heavy loads like a car.

6. Explain the principle of a hydraulic brake with a simple labelled diagram.
Step 1: (Diagram conceptually shows a foot pedal linked to a master cylinder filled with brake fluid, with brake lines branching out to wheel cylinders which push brake shoes against wheel rims).
Step 2: Principle: It operates using Pascal's law to safely stop vehicles.
Step 3: When the driver steps on the foot pedal, a small push is applied to a small piston in the master cylinder.
Step 4: This action creates pressure in the brake fluid.
Step 5: The pressure instantly travels through the brake pipes to the wheel cylinders at each wheel.
Step 6: The pressure acts on slightly larger pistons in the wheel cylinders, forcing them outward to press brake shoes against the rotating wheels, halting the car smoothly.

(E) NUMERICALS :

1. A hammer exerts a force of 1.5 N on each of the two nails A and B. The area of cross section of tip of nail A is 2 mm² while that of nail B is 6 mm². Calculate pressure on each nail in pascal.
Step 1: Write down the applied force for both nails: F = 1.5 N.
Step 2: Write the tip area for nail A: A₁ = 2 mm².
Step 3: Convert the area of nail A to square meters (m²) by multiplying by 10^-6: A₁ = 2 × 10^-6 m².
Step 4: Use the formula Pressure = Force / Area.
Step 5: Calculate the pressure on nail A: P_A = 1.5 / (2 × 10^-6) = 7.5 × 10⁵ Pa.
Step 6: Write the tip area for nail B: A₂ = 6 mm².
Step 7: Convert the area of nail B to square meters: A₂ = 6 × 10^-6 m².
Step 8: Calculate the pressure on nail B: P_B = 1.5 / (6 × 10^-6) = 2.5 × 10⁵ Pa.

2. A block of iron of mass 7.5 kg and of dimensions 12 cm × 8 cm × 10 cm is kept on a table top on its base of side 12 cm × 8 cm. Calculate :
(a) thrust and
(b) pressure exerted on the table top. Take 1 kgf = 10 N.
Step 1: (a) Identify the mass of the block: mass = 7.5 kg.
Step 2: Thrust is the weight of the block acting downwards. In kgf, thrust = 7.5 kgf.
Step 3: Convert kgf to Newtons by multiplying by 10: Thrust = 7.5 × 10 = 75 N.
Step 4: (b) Write down the dimensions of the base resting on the table: length = 12 cm, breadth = 8 cm.
Step 5: Calculate the area of the base in cm²: Area = 12 × 8 = 96 cm².
Step 6: Convert the area to standard m²: 96 / 10000 = 0.0096 m².
Step 7: Use the formula: Pressure = Thrust / Area.
Step 8: Calculate the pressure: P = 75 / 0.0096 = 7812.5 Pa.

3. A vessel contains water up to a height of 1.5 m. Taking the density of water 10³ kg m^-3, acceleration due to gravity 9.8 m s^-2 and area of base of vessel 100 cm², calculate :
(a) the pressure and
(b) the thrust, at the base of vessel.
Step 1: Write down the known values: depth h = 1.5 m, density ρ = 1000 kg m^-3, gravity g = 9.8 m s^-2.
Step 2: (a) Use the liquid pressure formula: Pressure = h × ρ × g.
Step 3: Calculate the pressure: P = 1.5 × 1000 × 9.8 = 14,700 Pa (which can be written as 1.47 × 10⁴ Pa).
Step 4: (b) Write down the base area: Area = 100 cm².
Step 5: Convert area to square meters: Area = 100 / 10000 = 0.01 m².
Step 6: Rearrange the pressure formula to find thrust: Thrust = Pressure × Area.
Step 7: Calculate the thrust: Thrust = 14700 × 0.01 = 147 N.

4. The area of base of a cylindrical vessel is 300 cm². Water (density = 1000 kg m^-3) is poured into it up to a depth of 6 cm. Calculate : (a) the pressure and (b) the thrust of water on the base. (g = 10 m s^-2).
Step 1: Write down the depth of water in cm: h = 6 cm.
Step 2: Convert depth to meters: h = 6 / 100 = 0.06 m.
Step 3: Write known constants: ρ = 1000 kg m^-3, g = 10 m s^-2.
Step 4: (a) Calculate liquid pressure using P = hρg: P = 0.06 × 1000 × 10 = 600 Pa.
Step 5: (b) Write down the area in cm²: Area = 300 cm².
Step 6: Convert area to meters squared: Area = 300 / 10000 = 0.03 m².
Step 7: Calculate thrust using Thrust = Pressure × Area: Thrust = 600 × 0.03 = 18 N.

5. (a) Calculate the height of a water column which will exert on its base the same pressure as the 70 cm column of mercury. Density of mercury is 13.6 g cm^-3.
(b) Will the height of the water column in part (a) change if the cross section of the water column is made wider ?
Step 1: (a) Set the pressure of the water column equal to the pressure of the mercury column.
Step 2: This means (height of water × density of water × g) = (height of mercury × density of mercury × g).
Step 3: Gravity (g) cancels out on both sides, leaving: h_w × ρ_w = h_m × ρ_m.
Step 4: Plug in known values: h_w × 1 g cm^-3 = 70 cm × 13.6 g cm^-3.
Step 5: Calculate the height of water: h_w = 952 cm.
Step 6: Convert to meters: h_w = 9.52 m.
Step 7: (b) Liquid pressure only depends on depth and density, not the container's shape or area.
Step 8: Therefore, No, the height will not change if the column is made wider.

6. The pressure of water on the ground floor is 40,000 Pa and on the first floor is 10,000 Pa. Find the height of the first floor.
(Take : density of water = 1000 kg m^-3, g = 10 m s^-2)
Step 1: Write down the pressure at the ground floor: P₁ = 40,000 Pa.
Step 2: Write down the pressure at the first floor: P₂ = 10,000 Pa.
Step 3: Find the difference in pressure between the two floors: ΔP = 40,000 - 10,000 = 30,000 Pa.
Step 4: This difference is purely due to the height (h) of the water column between the floors.
Step 5: Use the formula for liquid pressure difference: ΔP = h × ρ × g.
Step 6: Substitute the known values: 30,000 = h × 1000 × 10.
Step 7: Simplify the right side: 30,000 = h × 10,000.
Step 8: Solve for height h: h = 30,000 / 10,000 = 3 m.

7. A simple U tube contains mercury to the same level in both of its arms. If water is poured to a height of 13.6 cm in one arm, how much will be the rise in mercury level in the other arm ?
Given : density of mercury = 13.6 × 10³ kg m^-3 and density of water = 10³ kg m^-3.
Step 1: The water poured in creates a downward pressure in one arm.
Step 2: Calculate the pressure from the 13.6 cm water column: P_w = h_w × ρ_w × g.
Step 3: This pushes the mercury down in that arm by distance 'x', causing it to rise by 'x' in the other arm.
Step 4: The total height difference in the mercury levels is therefore 2x.
Step 5: Balance the pressures at the bottom level: Pressure of water column = Pressure of 2x mercury column.
Step 6: Set up the equation: 13.6 cm × (10³) × g = (2x) × (13.6 × 10³) × g.
Step 7: Cancel 'g' and '10³' from both sides: 13.6 = 2x × 13.6.
Step 8: Divide by 13.6: 1 = 2x.
Step 9: Solve for the rise x: x = 0.5 cm.

8. In a hydraulic machine, a force of 2 N is applied on the piston of area of cross section 10 cm². What force is obtained on its piston of area of cross section 100 cm² ?
Step 1: Write down Pascal's principle equation for a hydraulic machine: F₁ / A₁ = F₂ / A₂.
Step 2: Identify the known input values: F₁ = 2 N, A₁ = 10 cm².
Step 3: Identify the known output area: A₂ = 100 cm².
Step 4: Substitute the values into the equation: 2 / 10 = F₂ / 100.
Step 5: Simplify the left side: 0.2 = F₂ / 100.
Step 6: Multiply both sides by 100 to find the output force: F₂ = 0.2 × 100 = 20 N.

9. What should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of its brake shoe by exerting a force of 0.5 N on the pedal ?
Step 1: Write the given input force on the pedal (master cylinder): F₁ = 0.5 N.
Step 2: Write the desired output force at the brake shoe (wheel cylinder): F₂ = 15 N.
Step 3: Recall Pascal's principle formula: F₁ / A₁ = F₂ / A₂.
Step 4: Rearrange this formula to find the ratio of the areas (A₁ / A₂): A₁ / A₂ = F₁ / F₂.
Step 5: Substitute the known force values: Ratio = 0.5 / 15.
Step 6: Simplify the fraction by multiplying top and bottom by 10: 5 / 150.
Step 7: Divide top and bottom by 5 to get the final ratio: 1 / 30, or 1 : 30.

10. The areas of pistons in a hydraulic machine are 5 cm² and 625 cm². What force on the smaller piston will support a load of 1250 N on the larger piston ? State any assumption which you make in your calculation.
Step 1: Identify the area of the smaller piston: A₁ = 5 cm².
Step 2: Identify the area of the larger piston: A₂ = 625 cm².
Step 3: Identify the load (force) on the larger piston: F₂ = 1250 N.
Step 4: Use Pascal's formula: F₁ / A₁ = F₂ / A₂.
Step 5: Substitute the values: F₁ / 5 = 1250 / 625.
Step 6: Simplify the right side (1250 divided by 625 is exactly 2): F₁ / 5 = 2.
Step 7: Multiply by 5 to find the required force: F₁ = 2 × 5 = 10 N.
Step 8: State the assumption made: We assume there is no friction in the system and no leakage of liquid.

11. (a) The diameter of neck and bottom of a bottle are 2 cm and 10 cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2 kgf, what force is exerted on the bottom of the bottle ?
(b) Name the law/principle you have used to find the force in part (a)
Step 1: (a) Find the radius of the neck: r₁ = 2 cm / 2 = 1 cm.
Step 2: Calculate the area of the neck: A₁ = π × (1)².
Step 3: Find the radius of the bottom: r₂ = 10 cm / 2 = 5 cm.
Step 4: Calculate the area of the bottom: A₂ = π × (5)².
Step 5: Find the ratio of the areas (A₂ / A₁): 25π / 1π = 25.
Step 6: Write down the force applied to the neck: F₁ = 1.2 kgf.
Step 7: Use Pascal's law derived formula: F₂ = F₁ × (A₂ / A₁).
Step 8: Calculate the output force on the bottom: F₂ = 1.2 × 25 = 30 kgf.
Step 9: (b) The principle used to solve this is Pascal's law.

12. A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.
Step 1: Write down the applied force on the small piston: F₁ = 50 kgf.
Step 2: Write down diameter 1 (d₁) = 5 cm, and diameter 2 (d₂) = 25 cm.
Step 3: The ratio of areas is proportional to the square of the diameters: A₂ / A₁ = (d₂ / d₁)².
Step 4: Calculate the area ratio: (25 / 5)² = (5)² = 25.
Step 5: This means the large piston is 25 times larger in area than the small piston.
Step 6: Use the force multiplier formula: F₂ = F₁ × (A₂ / A₁).
Step 7: Calculate the final force: F₂ = 50 × 25 = 1250 kgf.

13. Two cylindrical vessels fitted with pistons A and B of area of cross section 8 cm² and 320 cm² respectively, are joined at their bottom by a tube and they are completely filled with water. When a mass of 4 kg is placed on piston A, find :
(i) the pressure on piston A,
(ii) the pressure on piston B, and
(iii) the thrust on piston B.
Step 1: (i) Write down the mass acting on piston A: mass = 4 kg, which means force F_A = 4 kgf.
Step 2: Write down the area of piston A: Area = 8 cm².
Step 3: Calculate pressure on A using P = F / Area: P = 4 / 8 = 0.5 kgf cm^-2.
Step 4: (ii) According to Pascal's law, pressure is transmitted equally throughout the fluid.
Step 5: Therefore, the pressure on piston B is exactly the same: 0.5 kgf cm^-2.
Step 6: (iii) Write down the area of piston B: Area_B = 320 cm².
Step 7: Rearrange the pressure formula to find thrust: Thrust = Pressure × Area.
Step 8: Calculate thrust on B: Thrust = 0.5 × 320 = 160 kgf.

14. What force is applied on a piston of area of cross section 2 cm² to obtain a force 150 N on the piston of area of cross section 12 cm² in a hydraulic machine ?
Step 1: Write down the known area of the small piston: A₁ = 2 cm².
Step 2: Write down the target output force: F₂ = 150 N.
Step 3: Write down the area of the large piston: A₂ = 12 cm².
Step 4: State Pascal's law formula: F₁ / A₁ = F₂ / A₂.
Step 5: Substitute the given values: F₁ / 2 = 150 / 12.
Step 6: Isolate F₁ by multiplying both sides by 2: F₁ = (150 × 2) / 12.
Step 7: Simplify the numerator: F₁ = 300 / 12.
Step 8: Divide to find the final force: F₁ = 25 N.

(F) ASSERTION - REASON TYPE QUESTIONS :
(Choose the answer based on the codes given below.)
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are True and R is not the correct explanation of A
(c) Assertion is false but reason is true
(d) Assertion is true but reason is false

(i) Assertion (A) : A brick exerts maximum pressure on ground when it is placed with its longest side vertical.
Reason (R) : Larger the area on which a given thrust acts, lesser is the pressure exerted by it.
Answer: (a) Both A and R are true and R is the correct explanation of A.
Step 1: Analyze the assertion: When the longest side is vertical, the base touching the ground has the smallest possible area. Smallest area gives maximum pressure, so assertion is true.
Step 2: Analyze the reason: Pressure = Thrust/Area. This confirms larger area gives lesser pressure, so reason is true.
Step 3: The inverse relationship explained in the reason is exactly why the assertion works, so R correctly explains A.

(ii) Assertion (A) : The walls of a dam are made thicker at the bottom.
Reason (R) : Pressure is the same in all directions about a point inside the liquid.
Answer: (b) Both A and R are True and R is not the correct explanation of A.
Step 1: Analyze the assertion: Dam walls are indeed thicker at the bottom, so A is true.
Step 2: Analyze the reason: Pascal's law states pressure is same in all directions at a point, so R is technically true.
Step 3: Determine the link: Dams are thicker at the bottom because pressure *increases with depth*, NOT just because it acts in all directions. Thus, R does not correctly explain A.

(iii) Assertion (A) : A hydraulic jack is used for squeezing oil out of linseed and cotton seeds.
Reason (R) : It works on the Pascal's principle.
Answer: (a) Both A and R are true and R is the correct explanation of A.
Step 1: Analyze the assertion: Hydraulic machinery (like presses) is indeed used industrially to extract oil by applying massive pressure, so A is true.
Step 2: Analyze the reason: Hydraulic machines work precisely because of Pascal's principle of pressure transmission, so R is true.
Step 3: The application of this immense pressure is directly due to the force multiplication of Pascal's principle, so R correctly explains A.

(G) CASE STUDY BASED QUESTION :
1. A hydraulic jack is used in a car workshop to lift a car. The jack consists of two connected pistons A (area = 5 cm²) and B (area = 250 cm²), filled completely with an incompressible hydraulic fluid (such as hydraulic oil). The fluid transmits pressure uniformly throughout the system. A force of 120 N is applied vertically downward on piston A.
Assuming the system to be ideal, answer the following questions:

(a) Calculate the pressure produced in the liquid by piston A.
Step 1: Write down the applied force on piston A: F = 120 N.
Step 2: Write down the area of piston A: Area = 5 cm².
Step 3: Convert the area into standard SI units (m²) by multiplying by 10^-4: Area = 5 × 10^-4 m².
Step 4: State the formula for pressure: Pressure = Force / Area.
Step 5: Calculate the pressure: P = 120 / (5 × 10^-4) = 2.4 × 10⁵ Pa.

(b) Determine the force exerted on piston B. Name the law used.
Step 1: Recall Pascal's law, which states that pressure is transmitted equally throughout an enclosed fluid.
Step 2: Because of this law, the pressure on piston B is exactly the same: 2.4 × 10⁵ Pa.
Step 3: Write down the area of piston B: Area = 250 cm².
Step 4: Convert area to m²: Area = 250 × 10^-4 m².
Step 5: Use the formula Force = Pressure × Area.
Step 6: Calculate the force: F = (2.4 × 10⁵) × (250 × 10^-4) = 6000 N.
Step 7: Name the law: The law used is Pascal's law.

(c) Does this hydraulic jack provide a gain in force or a gain in distance? Justify your answer.
Step 1: Look at the small input effort applied: 120 N.
Step 2: Look at the massive output force generated: 6000 N.
Step 3: Because a very small force is being multiplied into a much larger force, the machine acts as a force multiplier.
Step 4: Therefore, it provides a gain in force.

(d) If piston B rises by a height of 2 cm, calculate the distance moved by piston A.
Step 1: Understand that hydraulic oil is incompressible, meaning the volume of liquid pushed down by A exactly equals the volume pushed up at B.
Step 2: State the volume equation: Area of A × distance moved by A = Area of B × distance moved by B.
Step 3: Substitute the known values (keeping units in cm is fine here): 5 × distance of A = 250 × 2.
Step 4: Simplify the right side: 5 × distance of A = 500.
Step 5: Divide to find the distance: distance moved by A = 500 / 5 = 100 cm.

(e) What would happen if a small air bubble enters the liquid of the hydraulic jack?
Step 1: Remember the difference between gases and liquids: liquids cannot be compressed, but gases (like air bubbles) can be easily compressed.
Step 2: If an air bubble gets into the system, applying a downward force on piston A will first squeeze and compress that bubble.
Step 3: The pressure is absorbed by the bubble instead of being instantly transmitted to piston B.
Step 4: As a result, the hydraulic jack will lose efficiency and struggle to lift the car properly.